Skip to main content

Leibnitz theorem: y = e^(a tan⁻¹x)

Short answer

Leibnitz’s theorem: (uv)ₙ = Σ_{r=0}^n C(n,r) u_{n−r} v_r. For y = e^{a tan⁻¹x}, differentiate to get (1+x²)y₁ = a y. Applying the theorem (n+1) times yields (1+x²)y_{n+2} + (2nx+2x−a)y_{n+1} + n(n+1)y_n = 0.

State Leibnitz's theorem on higher order derivatives. If y = e^(a tan⁻¹x), prove that (1+x²)y_{n+2} + (2nx+2x−a)y_{n+1} + n(n+1)y_n = 0.

Leibnitz's theorem. If u and v are functions of x with nth derivatives, then the nth derivative of the product is

(uv)ₙ = uₙv + C(n,1)uₙ₋₁v₁ + C(n,2)uₙ₋₂v₂ + ... + u vₙ

where C(n,r) = n!/(r!(n−r)!).

Proof

y = e^(a tan⁻¹x). Then

y₁ = e^(a tan⁻¹x) · a/(1+x²) = a y/(1+x²)

(1 + x²) y₁ = a y …(1)

Differentiate (1) (n+1) times by Leibnitz's theorem. Take u = y₁ (or y), v = 1 + x² (v₁ = 2x, v₂ = 2):

(1+x²)yₙ₊₂ + (n+1)(2x)yₙ₊₁ + [(n+1)n/2](2)yₙ = a yₙ₊₁

That is

(1 + x²) yₙ₊₂ + (2nx + 2x − a) yₙ₊₁ + n(n+1) yₙ = 0. Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics I chapter 1 questions →